Some DNA sequences exist in circular forms as in the following figure, which shows a circular sequence CGAGTCAGCT", that is, the last symbolT" in CGAGTCAGCT" is connected to the first symbolC". We always read a circular sequence in the clockwise direction.

\epsfbox{p3225.eps}
Since it is not easy to store a circular sequence in a computer as it is, we decided to store it as a linear sequence. However, there can be many linear sequences that are obtained from a circular sequence by cutting any place of the circular sequence. Hence, we also decided to store the linear sequence that is lexicographically smallest among all linear sequences that can be obtained from a circular sequence.

Your task is to find the lexicographically smallest sequence from a given circular sequence. For the example in the figure, the lexicographically smallest sequence is ``AGCTCGAGTC". If there are two or more linear sequences that are lexicographically smallest, you are to find any one of them (in fact, they are the same).

Input

The input consists of T test cases. The number of test cases T is given on the first line of the input file. Each test case takes one line containing a circular sequence that is written as an arbitrary linear sequence. Since the circular sequences are DNA sequences, only four symbols, A, C, G and T, are allowed. Each sequence has length at least 2 and at most 100.

Output

Print exactly one line for each test case. The line is to contain the lexicographically smallest sequence for the test case.

The following shows sample input and output for two test cases.

Sample Input

2

CGAGTCAGCT

CTCC
Sample Output

AGCTCGAGTC
CCCT

我的代码如下

#include <iostream>
#include <cstring>
using namespace std;
int small(const char* s,int p,int q);
#define maxn 105
int main()
{
    int n;//测试数
    cin>>n;
    while(n--){
        //为何不能用char *str={0};？
        char str[maxn];
        cin>>str;
        int len=strlen(str);
        int ans=0;
        for(int i=1;i<len;i++){
            if(small(str,i,ans))
                ans=i;
        }
        for(int i=0;i<len;i++){
            putchar(str[(ans+i)%len]);
        }
        putchar('\n');
    }
    return 0;

}

int small(const char* s,int p,int q){  //比较表示法p和表示法q字典序大小
    int n=strlen(s);
    for(int i=0;i<n;i++){
        if(s[(p+i)%n]!=s[(q+i)%n])
            return s[(p+i)%n]<s[(q+i)%n];
    }
    return 0;


}

问题疑惑在注释里，如果用注释里的会报空指针错误（ returned -1073741819 (0xC0000005)）问的问题很小白，让各位见笑了囧。

demo_zzt

大一刚刚学习ACM 求大佬指点错误。

3条回答

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